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I recently ran into this interview question and am wondering if my solution is correct. The setting is

We have a bag containing one coin of each type, i.e. we have one $1$ cent coin, one $2$ cent coin, one $5$ cent coin, one $10$ cent coin, one $20$ cent coin, one $50$ cent coin, one $100$ cent coin and one $200$ cent coin. We random select three coins from this bag, what is the probability that the sum of the values of the coins is at least $80$ cents? Start off with giving an initial guess for this probability, without computing anything, and explain your estimate.

My solution: For the estimation part, I figured there were $\binom{7}{2}$ configurations of three coins that contain the $200$ cent coin. Furthermore, there's also $\binom{7}{2}$ configurations of three coins that contain the $100$ cent coin. This already sums to $21+21$ configurations. However, this sum contains some configurations twice, so it is an overestimation. In total there are $\binom{8}{3}=56$ configurations of three coins. Thus the probability of taking three coins with a total value of atleast $80$ cents, is approximately $\frac{42}{56}\approx\frac{2}{3}$ (rough estimation, I know).

For the exact solution. I know that there are $\binom{7}{2}$ configurations of three coins that contain the $200$ cent coin, thus automatically satisfying that the sum is above $80$ cents. Then, to avoid duplicate configurations, there are $\binom{6}{2}$ configurations of coins containing the $100$ cent coin. Then there is one last configuration of $50$ cents, $10$ cents and $20$ cents, that also makes $80$ cents. Thus there are $\binom{7}{2}+\binom{6}{2}+1=21+15+1=37$ configurations. Thus the probability of taking atleast $80$ cents when randomly grabbing $3$ coins, is $\frac{37}{56}$.

Is this correct? Any help is appreciated.

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  • $\begingroup$ You really should list the values of all the coins explicitly. Not everybody lives in the Euro zone! $\endgroup$ – TonyK Feb 16 '19 at 21:20
  • $\begingroup$ For anyone curious, the euro coin denominations are 1 cent, 2 cents, 5 cents, 10 cents, 20 cents, 50 cents, 1 euro and 2 euro. (And 1 euro is 100 cents of course.) $\endgroup$ – Minus One-Twelfth Feb 16 '19 at 21:23
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    $\begingroup$ So it was sheer laziness? $\endgroup$ – TonyK Feb 16 '19 at 21:37
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    $\begingroup$ @TonyK I edited the post. Hope you can feel relaxed once more. $\endgroup$ – Charlie Shuffler Feb 16 '19 at 21:41
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    $\begingroup$ @TonyK Fair enough, sorry. $\endgroup$ – Charlie Shuffler Feb 16 '19 at 22:00
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To getting at least 80 cents:

we have 1€ and two from the rest: ${7\choose 2} =21$

we don't have 1€, but we have 2€: ${6\choose 2} =15$

we don't have 1€ and not 2€, then it is not possible if we have 10,20 and 50 cents

So we have 37 good possibilites among ${8\choose 3} = 56$ so the probability is $$P= {37\over 56}$$

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    $\begingroup$ That is incorrect for multiple reasons, but also because there are $8$ different coins in total. $\endgroup$ – Charlie Shuffler Feb 16 '19 at 21:03
  • $\begingroup$ After the edit, that is also what I got as final answer. Thanks for the help! $\endgroup$ – Charlie Shuffler Feb 16 '19 at 21:10
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More elegant (IMO) and easily doable with paper and pen:

there are $~2~$ ways to get at least $~80c~$

$~1: 50+20+10 \to$ probability is $~3/8~$ for first draw, $~2/7~$ for $2$nd draw, $~1/6~$ for third draw for total of: $$(3*2*1)/(8*7*6)$$

$~2: 100 +~$ anything or $~ 200+$anything

which is $~1-(~$Probability of not getting $~100~$ or $~200)~$

Probability of not getting $~100~$ or $~200~$ is $~6/8~$ for first draw, $~5/8 ~$ for $2$nd draw and $~4/8~$ for $3$rd draw

so total probability for option $~2~$ is $$1-(6*5*4)/(8*7*6)$$

So total probability to get at least $~80c~$ is

$$3*2*1/(8*7*6)+1-(6*5*4)/(8*7*6)$$ Simplify the fractions by dividing out the $6:$ $$1/(8*7)+1-5*4/(8*7)$$ $$=1+1/56-20/56$$ $$=(56+1-20)/56$$ $$=37/56$$

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